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I am a SAT tutor and this problem comes up in our math lessons and quite frankly, baffles me as it seems very unintuitive.

If four workers can dig four ditches in four hours, how many hours will it take for two workers to dig two ditches?

The answer is 4 (at least according to the book) and I can't wrap my head around the way it is explained in the book (which unfortunately I don't have with me). They way I went about solving is very questionable and leaves me doubtful of whether or not it is indeed the correct answer.

I put in the 4 to keep the equation balanced with the initial conditions. Then for the question itself:

xy = 4k

Then plugged in values x = 2 and k = 2 by the question proposed.

2y = 4(2)

y = 4

This is the right answer however, like I said it doesn't feel right and I am uncertain that not only this isn't the correct way to do the problem but that the problem and answer itself is faulty. Any input/advice would be much appreciated.

Okay then how does one get to the conclusion that 2 workers can dig two ditches in four hours? Because if one can dig one in four hours, then adding another worker should effectively cut the time in half would it not (thinking in a real world scenario).

If four workers can dig four ditches in four hours, and each worker is doing 1/4th of the work, then one worker would need to do four times the work to dig the same number of ditches. So one worker can dig four ditches in (4)(4)=16 hours. But if he only needs to do one fourth of the work, then he can dig one ditch in (16/4) = 4 hours.
Formally, if n workers can dig d ditches in h hours, then one worker can dig one ditch in nh/d hours.
This is a good place to start. Figure out the base rate, then we can go from there.
If one worker can dig a ditch in four hours, then he can dig two ditches in (4)(2) = 8 hours. If you add another worker, then each only has to do half the time, so two workers can dig two ditches in (8/2) = 4 hours.
In general, if one worker can dig one ditch in t hours, then m workers can dig D ditches in (tm/D) hours.